diff --git a/sql/task1.sql b/sql/task1.sql
index 90de336ca..785a2a00b 100644
--- a/sql/task1.sql
+++ b/sql/task1.sql
@@ -1,14 +1,54 @@
 -- Problem 1: Retrieve all products in the Sports category
 -- Write an SQL query to retrieve all products in a specific category.
+-- Load data from files .csv files:
+
+SELECT 
+	product_id, product_name
+FROM 
+	products
+WHERE 
+	category_id = 8;  -- The category id for sports is 8, hence I ran the query by category ID. You can also query by category name if you use JOIN.
+	
 
 -- Problem 2: Retrieve the total number of orders for each user
 -- Write an SQL query to retrieve the total number of orders for each user.
 -- The result should include the user ID, username, and the total number of orders.
 
+SELECT 
+	users.user_id, users.username, COUNT(order_id) AS total_orders
+FROM 
+	orders
+JOIN 
+	users ON orders.user_id = users.user_id
+GROUP BY 
+	users.user_id, users.username; -- This retrieves the total number of ORDERS, not the total amount of items ordered.
+
 -- Problem 3: Retrieve the average rating for each product
 -- Write an SQL query to retrieve the average rating for each product.
 -- The result should include the product ID, product name, and the average rating.
 
+SELECT 
+	products.product_id, products.product_name, AVG(rating) AS average_rating
+FROM 
+	reviews
+JOIN 
+	products ON products.product_id = reviews.product_id -- Ensures that we get the average for each product
+GROUP BY 
+	products.product_id;
+
+
 -- Problem 4: Retrieve the top 5 users with the highest total amount spent on orders
 -- Write an SQL query to retrieve the top 5 users with the highest total amount spent on orders.
 -- The result should include the user ID, username, and the total amount spent.
+
+SELECT 
+	users.user_id, username, SUM(total_amount) AS total_spent
+FROM 
+	users, orders
+WHERE 
+	users.user_id = orders.user_id 
+GROUP BY 
+	users.user_id 
+ORDER BY 
+	total_spent DESC
+LIMIT 5; -- Only retrieve the top 5 users
diff --git a/sql/task2.sql b/sql/task2.sql
index ad2596731..fd2fb204f 100644
--- a/sql/task2.sql
+++ b/sql/task2.sql
@@ -3,17 +3,92 @@
 -- The result should include the product ID, product name, and the average rating.
 -- Hint: You may need to use subqueries or common table expressions (CTEs) to solve this problem.
 
+WITH AvgRating AS (
+    SELECT 
+        products.product_id, 
+        products.product_name, 
+        AVG(rating) AS product_average_rating
+    FROM 
+        reviews
+    JOIN 
+        products ON products.product_id = reviews.product_id
+    GROUP BY 
+        products.product_id
+) -- This block of code creates the CTE
+
+SELECT 
+    product_id, 
+    product_name, 
+    product_average_rating
+FROM 
+    AvgRating
+WHERE 
+    product_average_rating = (SELECT MAX(product_average_rating) FROM AvgRating); -- get the max value from AvgRating CTE
+
 -- Problem 6: Retrieve the users who have made at least one order in each category
 -- Write an SQL query to retrieve the users who have made at least one order in each category.
 -- The result should include the user ID and username.
 -- Hint: You may need to use subqueries or joins to solve this problem.
 
+SELECT 
+    username
+FROM 
+    users
+WHERE (
+    SELECT 
+        COUNT(DISTINCT category_id) 
+    FROM 
+        orders 
+    JOIN 
+        order_items ON order_items.order_id = orders.order_id 
+    JOIN 
+        products ON order_items.product_id = products.product_id
+    WHERE orders.user_id = users.user_id
+    )  = 
+    
+    (SELECT 
+        COUNT(DISTINCT category_id) -- check if the user has made at least one order in each category
+     FROM 
+        categories); 
+        
+	-- There are no users that have made at least one order in each category!
+
 -- Problem 7: Retrieve the products that have not received any reviews
 -- Write an SQL query to retrieve the products that have not received any reviews.
 -- The result should include the product ID and product name.
 -- Hint: You may need to use subqueries or left joins to solve this problem.
 
+SELECT 
+    product_id, product_name
+FROM 
+    products
+WHERE 
+    product_id NOT IN (
+        SELECT 
+            product_id
+        FROM 
+            reviews
+    ); -- Used a subquery to retrieve the products without reviews
+
+-- There are no products without reviews!
+    
 -- Problem 8: Retrieve the users who have made consecutive orders on consecutive days
 -- Write an SQL query to retrieve the users who have made consecutive orders on consecutive days.
 -- The result should include the user ID and username.
--- Hint: You may need to use subqueries or window functions to solve this problem.
\ No newline at end of file
+-- Hint: You may need to use subqueries or window functions to solve this problem.
+
+SELECT DISTINCT 
+	ub.user_id, 
+    ub.username
+FROM (
+	SELECT 
+		u.user_id, 
+        u.username,
+		datediff(o.order_date, LAG(o.order_date, 1) OVER (PARTITION BY u.user_id ORDER BY o.order_date)) AS difference
+	FROM
+		users u, orders o
+	WHERE 
+		u.user_id = o.user_id 
+) ub
+WHERE 
+	ub.difference is not null and ub.difference = 1 -- return if it was a consecutive day
\ No newline at end of file
diff --git a/sql/task3.sql b/sql/task3.sql
index f078a9439..43df54d89 100644
--- a/sql/task3.sql
+++ b/sql/task3.sql
@@ -3,17 +3,101 @@
 -- The result should include the category ID, category name, and the total sales amount.
 -- Hint: You may need to use subqueries, joins, and aggregate functions to solve this problem.
 
+
+SELECT 
+	categories.category_id, category_name, SUM(quantity * unit_price) AS total_amount
+FROM 
+	categories
+JOIN 
+	products ON categories.category_id = products.category_id
+JOIN 
+	order_items ON products.product_id = order_items.product_id
+GROUP BY 
+	categories.category_id
+ORDER BY 
+	total_amount DESC
+LIMIT 3;
+
 -- Problem 10: Retrieve the users who have placed orders for all products in the Toys & Games
 -- Write an SQL query to retrieve the users who have placed orders for all products in the Toys & Games
 -- The result should include the user ID and username.
 -- Hint: You may need to use subqueries, joins, and aggregate functions to solve this problem.
 
+SELECT
+	user_id,
+    username
+FROM 
+	users
+WHERE (
+    SELECT COUNT(DISTINCT order_items.product_id) 
+    FROM 
+		order_items
+    JOIN 
+		orders ON orders.order_id = order_items.order_id 
+    JOIN 
+		products ON products.product_id = order_items.product_id
+    JOIN 
+		categories ON products.category_id = categories.category_id
+    WHERE orders.user_id = users.user_id and category_name = 'Toys & Games') 
+    = 
+    (SELECT COUNT(Distinct product_id) 
+     FROM 
+		products
+     JOIN 
+		categories ON categories.category_id = products.category_id
+     WHERE 
+		category_name = 'Toys & Games');
+        
+	-- Sarah placed the most orders for Toys & Games
+
 -- Problem 11: Retrieve the products that have the highest price within each category
 -- Write an SQL query to retrieve the products that have the highest price within each category.
 -- The result should include the product ID, product name, category ID, and price.
 -- Hint: You may need to use subqueries, joins, and window functions to solve this problem.
 
+
+WITH MaxP AS (
+    SELECT
+		category_id, MAX(price) as highest_price
+    FROM 
+		products
+    GROUP BY 
+		category_id
+)
+
+SELECT 
+	product_id, 
+    product_name, 
+    products.category_id, 
+    highest_price
+FROM 
+	products
+JOIN 
+	MaxP ON products.category_id = MaxP.category_id
+WHERE
+	price = highest_price;
+
 -- Problem 12: Retrieve the users who have placed orders on consecutive days for at least 3 days
 -- Write an SQL query to retrieve the users who have placed orders on consecutive days for at least 3 days.
 -- The result should include the user ID and username.
--- Hint: You may need to use subqueries, joins, and window functions to solve this problem.
+-- Hint: You may need to use subqueries, joins, and window functions to solve this problem
+
+WITH OrderDateRank AS ( 
+    SELECT 
+        ordered.user_id,
+        ordered.order_date,
+        RANK() OVER (PARTITION BY ordered.user_id ORDER BY ordered.order_date) AS ranked
+    FROM Orders ordered
+)
+
+SELECT 
+    users.user_id,
+    users.username
+FROM 
+	OrderDateRank ordered
+INNER JOIN 
+	Users users ON ordered.user_id = users.user_id
+WHERE 
+	ordered.ranked >= 3;
+    
+    -- There are no users who retrieved orders on consecutive days
\ No newline at end of file
diff --git a/tests/test_sql_queries.py b/tests/test_sql_queries.py
index 22b25d546..6e45799d0 100644
--- a/tests/test_sql_queries.py
+++ b/tests/test_sql_queries.py
@@ -6,11 +6,11 @@ class TestSQLQueries(unittest.TestCase):
     def setUp(self):
         # Establish a connection to your test database
         self.conn = psycopg2.connect(
-            dbname='your_dbname',
-            user='your_username',
-            password='your_password',
-            host='your_host',
-            port='your_port'
+            dbname='test_db',
+            user='alinco',
+            password='123',
+            host='localhost',
+            port='3306'
         )
         self.cur = self.conn.cursor()
 
@@ -52,4 +52,4 @@ def test_task2(self):
     # Add more test methods for additional SQL tasks
 
 if __name__ == '__main__':
-    unittest.main()
\ No newline at end of file
+    unittest.main()